1925.]
SUSPENSION BRIDGE CATENARY
425
A GENERAL FORM
OF THE SUSPENSION BRIDGE CATENARY
BY IRA FREEMAN
1. Introduction. The form of the curve assumed by a
uniform chain or string hanging freely between two supports was first investigated by Galileo, who erroneously
determined it to be a parabola; Jungius detected Galileo's
error, but the true form was not discovered until 1691,
when James Bernoulli published it as a problem in the
ACTA ERUDITORUM. He also considered the cases when the
chain was (1) of variable density, (2) extensible, (3) acted
upon at each point by a force directed to a fixed center.
These curves attracted much
attention, and were discussed by
John Bernoulli, Leibnitz, Huygens,
David Gregory and others. A review of the literature shows, however, that in the problem of the
inextensible string acted upon by
gravitational forces, only two
special cases are considered:—the
case where there is simply a string
of uni-form linear density, and the
case of a string of inappreciable °v
x
weight supporting a uniform horizontal load. Moreover, the modern text-books in the subject appear to shun a more general case, and content
themselves with presenting only the two very special cases
mentioned.
The object of the present paper is to consider what
happens when a cable of appreciable uniform linear density
supports a uniform horizontal load, conditions that are
very nearly realized in the suspension-bridge. The texts
solve the special cases by the solution of two first-order
426
IRA FREEMAN
[October,
differential equations, the first of which gives the intrinsic
equation of the curve. The following method uses one
second-order differential equation, and does not involve
the intrinsic equation as an intermediate step.
2. Densities Different and not Zero. Both the cable and
the bridge platform have uniform linear densities ivhich are
appreciable, and, in general, different.
Let a be the weight per unit length of the cable, and
let fi represent that of the bridge platform. If any arc s
is in equilibrium, the horizontal component of the tensions,
at the extremities of the arc, will be the same constant
which we may call T. The difference of the vertical components of the extreme tensions will equal the total weight
on the arc; that is, we shall have
T(tan 0 2 —tan 0i) =
as-{-fix.
If s is small, this becomes
,àly
(1)
T^ax = «(A/JU
£ + *,
since
tan 0 2 —tan di =
-r-\-~-\dx.
dx\dxi
But since
ds_
dx
we may write
^dx- • V ' +J ( f )\dx]
'+'2
If p = dyldx,
we have
dp
(2)
= £to
'
a vT+j?+6
where a = a/T and 6 = fi/T. The equation (2) is the
differential equation of the curve. To integrate it, put
p = tan 0 ; then
sec 2 0^0 _
& + &sec0
'
do
cos 6(a-\-bcos
_
0)
1925.]
SUSPENSION BRIDGE CATENARY
427
By the method of partial fractions, we may write
cos 0
Integrating, we find
(A)
Me
_ ^ _ _
(3)
-
adx
a + &cos0
log (sec 0 + tan 0)
b
- V¥^
1
log
[b +
acos6-\-Vb2—a2sin0
l
^+ï^e
J] =- «°
the constant of integration being zero if we let x — 0
when 0 = 0. Dividing (2) by dy, we may write
(
,
dp
dy
aVT^+b
dx
1
pdp
,
~ ~dy ~~~ 7' aVT+p+6 ~
To integrate this put, as before p = tan 0.
V
Then
2
tan 0 sec 0^0
.
sec 0[d(sec 0)]
7
—-—— = dy
or
= ay.
A , 7
y
a sec 0 + &
a sec 0 + &
Integrating, we find
& + a s e c 0 — b log (& + a sec 0) = aay + JK".
liy = y0 when 0 = 0, J5T= & + a — blog(a + b)—a2y0,
and hence
(B)
a(sec , - l ) + 6 log [
1 R
? ^ _ ] = «V~*).
The equations (A) and (B) are the parametric equations of
the curve, the parameter being 0, the angle of inclination.
3. Alternative Forms. If b<Ca, equation (A) in its
present form will involve square roots of negative quantities.
But in this case there is a form of (A) which is real, and
which is therefore better suited for computation. This
form will now be developed.
The second term of the left member of (A) is now
&__
VW^2
p + acosfl+l/& 2 — a 2 sin0l
°gL
a + 5cos0
J
'
428
IRA FREEMAN
[October,
that is,
b
iV^-b2
z =
.
I b + aco$d-\-iV a2—-& 2 sin0]
°g 1
a + &cos0
J'
or
^=^
or
cos I
b + acos6 +
iVa*-b*sme
a + b cos 6
eVa*— b2
. IzVa2
r
) + *3sin
m
\—I
b + a cos 0 + iVa2—&2 sin (9
a + & cos 0
Equating the imaginary parts,
2
sm
IzV^Ta —&*\
]
/gKfl
=
y^^sinö
a + & cos 0 '
= 2 sm x "Ka22—&
—b2
LL a + &cos0
so that (A) may be replaced by the equation
(C)
Iog(sec0 + tan0)
b
v\
2
2
. AVa
sin 01 =
JVa2—&
—usinai
-1
sm L a + frcos0 J = ax.
4. Bridge Density Zero. If the linear density of the
bridge becomes zero, the equations reduce to those of the
common catenary. Here b = 0, so (A) takes the simple
form log (sec 0 + tan 0) = ax, since the second term reduces to zero. Hence we may write
sec 0 = —(e?°*-\-e~ax).
z
The equation (B) becomes sec 0 — 1 = a(y—y0), since the
argument of the logarithm never becomes infinite. Combining
these results, we have
eax =
sec
(9 + l/sec 2 a — 1 ,
\{eax+e~™)
or
= l + a(y—yo),
which is the equation of the common catenary.
1925.]
429
SUSPENSION BRIDGE CATENARY
5. Cable Density Zero. When we put the linear density
of the cable equal to zero, a becomes zero, and this makes (A)
and (B) reduce to identities. We can, however, obtain the
equations ivhich should result. If we divide (A) throughout
by a and then put a = 0, the result is an indeterminate
form, of the type zero divided by zero. This is evaluated
by the rule of de L'Hospital:
{b/(a2-b2)}{[Vb~2
—a2 (b + a cos 0 + Vb2— a2 sin d)
— (a + b cos 0) (Vb2—a2 cos 0—a sin o)] ~
[{b + a cos 0 +VV:=ö?$me)(a
+ &cos 0)] +
2
\b-\-aco$0+Vb —a2sin0
L
a+6cos0
Vb2- -aa
•]}•
When we now put a
0, this reduces to (1/6) tan 0, and
(A) takes the form
(A1)
bx.
tan 0
log
Similarly, when we divide (B) through by a2 and take the
quotient of the derivatives, we have
(sec Ö — 1) 1
b2
(a + &)(& + « sec 0)
2a
a=0
Applying the rule a second time,
&(l + secfl) + 2flsecfl
y(sec0—1)
{a + b)2(b + asecd)2 a=a =
so that (B) takes the form
(B1)
1
tan 2 0
2b
y—yo-
Combining (A1) and (B1), we may write
y- -Vo =
jar
which is the equation of a parabola.
T H E UNIVERSITY OF CHICAGO
0^
0
2ÏUn2e>